i don't know what kind of logic/proofs you are studying, so i don't know how you want these set up, but here's the idea for this one.

clearly s or ~s is true. if s is true, then we have ~a & b. however, this is the negation of a or ~b (DeMorgan's law), and so we cannot have that. thus, ~s must be true. but ~s => c, so that we can conclude c

2) ~r
( b or c) -> r
~d or m
d -> ~ ( c or p)
_______________
p -> m

i don't think this is correct. are you sure there is no typo?

3) b -> c
~c
(~b or ~p) -> s
_______________
s V q

we must show either s or q is true, perhaps both. since we have b => c and ~c, then we have ~b. if ~b is true, then ~b or ~p is true. and hence, the last implication says s is true. if s is true, then s or q is true, and we conclude accordingly.